Question

using division show that (x+1) is a factor of (2x²+3x+1)​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm \: p(x) = {2x}^{2} + 3x + 1$

and

$\rm \: q(x) = x + 1$

Now, we have to show that q(x) is a factor of p(x) by using Division Method. So, we have to divide p(x) by q(x) using Long Division Method and in order to show that q(x) is a factor of p(x), we have to show that remainder is 0.

So, By using Long Division Method, we have

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{2x + 1}}}\\ {{\sf{x + 1}}}& {\sf{ {2x}^{2} + 3x + 1}} \\{\sf{}}&\underline{\sf{ { - 2x}^{2} - 2x \: \: \: \: \: \: \: \: \: \: }}\\{\sf{}}&{\sf{ \: \: \: \: \: \: \: \: \: \: \: \: x + 1}}\\{\sf{}}&\underline{\sf{ \: \: \: \: \: \: \: \: \: \: - x - 1}}\\{\sf{}}&\underline{\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: }}\end{array}\end{gathered}\end{gathered}\end{gathered}$

So,

$\rm\implies \: {2x}^{2} + 3x + 1 \: is \: divisible \: by \: x + 1$

$\rm\implies \: x + 1 \: is \: factor \: of \: {2x}^{2} + 3 x + 1$

$\rule{190pt}{2pt}$

Additional Information :-

1. Remainder Theorem :- This theorem states that if a polynomial p(x) of degree greater than or equals to 1 is divided by x - a, then remainder is p(a).

2. Factor Theorem :- This theorem states that if a polynomial p(x) of degree greater than or equals to 1 is divisible by x - a, then p(a) = 0.

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

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